You have found the following ages (in years) of all 4 snakes at your local zoo: $ 5,\enspace 5,\enspace 7,\enspace 20$ What is the average age of the snakes at your zoo? What is the variance? You may round your answers to the nearest tenth.
Because we have data for all 4 snakes at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{5 + 5 + 7 + 20}{{4}} = {9.3\text{ years old}} $ Find the squared deviations from the mean for each snake. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $5$ years $-4.3$ years $18.49$ years $^2$ $5$ years $-4.3$ years $18.49$ years $^2$ $7$ years $-2.3$ years $5.29$ years $^2$ $20$ years $10.7$ years $114.49$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{18.49} + {18.49} + {5.29} + {114.49}} {{4}} $ $ {\sigma^2} = \dfrac{{156.76}}{{4}} = {39.19\text{ years}^2} $ The average snake at the zoo is 9.3 years old. The population variance is 39.19 years $^2$.